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As an example, for 17, being reduced modulo 7 is 17mod7 = 3, and for -3 modulo 5 is -3mod5 = 2. Given a modulo, M, and two integers, A and B, with modular representations r and s, respectively, we have: A r ¼Cþ M M B s s ¼ BmodM ) ¼ D þ M M r ¼ AmodM ) For the modular representation of the sum of A and B (A ? B)modM, we have: AþB rþs ¼ ðC þ D Þ þ M M Taking into account that r ? s may be greater than M, from the last equation it results: ðA þ BÞmodM ¼ ðAmodM þ BmodM ÞmodM For the modular representation of the product aA it results: aA ar ¼ aC þ M M Again, because ar may be greater than M, from this equality it results: aAmodM ¼ faðAmodM ÞgmodM Applying all of these expressions to the development of an integer number N in positional notation as a sum of powers of the radix, N ¼ an bn þ anÀ1 bnÀ1 þ Á Á Á þ a1 b1 þ a0 b0 12 1 Number Systems we have: ﬃ À Á NmodM ¼ an ðbn modM Þ þ anÀ1 bnÀ1 modM Á Á Á À Á À ÁÉ þ a1 b1 modM þ a0 b0 modM modM The modular reductions of the different powers of the radix can be pre-computed, being possible the simplification of the NmodM computing applying the following expression: n an b þ Á Á Á þ a0 b0 ¼ jan jbn jm þ Á Á Á þ a0 jb0 jm j m m Obviously, if M is a power of the base, M = bk, the modular reduction of N = anbn ?

In fact: B4 ¼ N1 À N2 þ D ¼ ðN1 þ DÞ À ðN2 þ DÞ þ D ¼ B1 À B2 þ D Thus, additions and subtraction must be implemented as different operations, and the biased is always involved, subtracting or adding it to the previous sum or subtraction. 4 Negative Integer Numbers 35 complementing the most significant bit. 4), this issues are shown: Thus, the addition/subtraction when using biased representation with D = 2m-1 can be implemented by means of a binary adder/subtractor of m bits (in SM), and complementing the most significant bit of the result.

5) ? (-8): comparing the magnitudes the result will be negative, with a magnitude of: 8 À 5 ¼ 8 þ "5 þ 1 ¼ 1000 þ 1010 þ 0001 ¼ 10011 that results 0011 if the carry is discarded. Joining the sign bit and the magnitude, the final result is 10011. Later will be shown this is equivalent to use one’s complement. The presented subtraction way only produces correct results if the minuend is greater than the subtrahend. In other words, for adding/subtracting with SM representation, a comparator is needed in order to decide which operand is the minuend and which the subtrahend when a subtraction is required.